【數論（擴充套件的歐幾里德）】ZOJ-3593-One Person Game
阿新 • • 發佈：2019-01-08
#include<iostream>
#include<cstdio>
#include<cmath>
using namespace std;
#define LL long long
struct euclid
{
LL x,y,d;
};
LL labs(LL x)
{
if(x<0)return -x;
return x;
}
euclid Euclid(LL a,LL b)
{
euclid eld1;
eld1.d=a;
eld1.x=1;
eld1.y=0;
if(!b)return eld1;
euclid eld2=Euclid(b,a%b);
eld1.d=eld2.d;
eld1.x=eld2.y;
eld1.y=eld2.x-a/b*eld2.y;
return eld1;
}
LL cal(LL a,LL b)
{
LL aa=labs(a),bb=labs(b);
if(a*b>0)return max(aa,bb); //當a，b同號時，就取它們絕對值之中大的那個數
return aa+bb; //異號時，就是兩個數絕對值的和
}
int main()
{
//freopen("a.txt","r",stdin);
int ca;
scanf("%d",&ca);
while(ca--)
{
LL s,t,a,b;
scanf("%lld%lld%lld%lld",&s,&t,&a,&b);
LL d=labs(s-t);
euclid eld=Euclid(a,b); //先用擴充套件的歐幾里德判斷是否有解
if(d%eld.d)
{
printf("-1\n");
continue;
}
eld.x*=d/eld.d;
eld.y*=d/eld.d;
LL k,x,y,step1,step2;
if(eld.x>eld.y)
{
k=(eld.x-eld.y)/(a/eld.d+b/eld.d);
x=eld.x-k*b/eld.d;
y=eld.y+k*a/eld.d; //求出當eld.x-k*（b/bel.d）=eld.y+k*（a/bel.d）的k值，因為當x==y時，x+y的值最小
step1=cal(x,y);
x-=b/eld.d;
y+=a/eld.d;
step2=cal(x,y); //求出k值左右的兩種情況
}
else
{
k=(eld.y-eld.x)/(a/eld.d+b/eld.d);
x=eld.x+k*b/eld.d;
y=eld.y-k*a/eld.d; //求出當eld.x+k*（b/bel.d）=eld.y-k*（a/bel.d）的k值
step1=cal(x,y);
x+=b/eld.d;
y-=a/eld.d;
step2=cal(x,y);
}
printf("%lld\n",min(step1,step2));
}
return 0;
}